Question: What is the missing constant term in the perfect square that starts with $x^2-16x$ ?
Explanation: Let $b$ be the missing constant term. Let's assume $x^2-16x+b$ is factored as the perfect square $(x+a)^2$. $\begin{aligned} (x+a)^2&=x^2+{2a}x+{a^2} \\\\ &=x^2{-16}x+ b \end{aligned}$ For the expressions to be the same, ${2a}$ must be equal to ${-16}$, and ${a^2}$ must be equal to $ b$. From ${2a=-16}$ we know that $a=-8$. Now, from ${a^2=b}$ we know that $b=(-8)^2=64$. Indeed, $x^2-16x+64$ is factored as $(x-8)^2$. In conclusion, the missing constant term in the perfect square that starts with $x^2-16x$ is $64$